A3 The diagram shows the structure of a metal. Key + metal ion – electron [Figure 6.1: A lattice of positive metal ions with a sea of delocalised electrons surrounding them.]
- (a)(i)Refer to this structure to explain why metals are malleable,[2]
- (a)(ii)metals conduct electricity.[1]
- (b)The table shows the ease with which different metal oxides can be reduced. metal oxide | ease of reduction calcium oxide | not reduced by carbon at 1800°C iron(III) oxide | reduced by carbon at 650 °C silver oxide | reduced by heating without carbon titanium(IV) oxide | reduced by carbon at 1800 °C but not at 650 °C Use the information in the table to place the metals calcium, iron, silver and titanium in order of their reactivity. least reactive most reactive[1]
- (c)(i)Iron(III) oxide can be reduced by carbon monoxide. Fe2O3 + 3CO → 2Fe + 3CO2 Calculate the maximum mass of iron that can be formed when 14.4g of iron(III) oxide is reduced by excess carbon monoxide. Give your answer to three significant figures. mass of iron =[3]
- (c)(ii)Calculate the maximum volume of carbon dioxide, in dm³, produced by this reaction, at room temperature and pressure. volume of carbon dioxide[2]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2017). The mark or mark type is shown in brackets.
- [1] A3(a)(i) reference to layers
- [1] A3(a)(i) (layers) slide
- [1] A3(a)(ii) mobile electrons / electrons can move
- [1] A3(b) silver < iron < titanium < calcium
- [1] A3(c)(i) mol Fe2O3 = 14.4 / 160 OR 0.090
- [1] A3(c)(i) mol Fe = 2 × 0.090 OR 0.180
- [1] A3(c)(i) mass = (0.180 × 56) = 10.1
- [1] A3(c)(ii) mol CO2 = 3/2 × 0.18 OR 0.27
- [1] A3(c)(ii) volume = 0.27 × 24 = 6.48 (dm³)
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